Question

In a survey of $500$ TV viewers: $285$ watch football (F), $195$ hockey (H), $115$ basketball (B); $45$ watch F&B, $70$ watch F&H, $50$ watch H&B, and $50$ watch none. How many watch exactly one of the three games? 

• A. 325
• B. 405
• C. 310
• D. 372

Hint:

For "exactly one", compute each group as:  
\[\text{Only }F = F - (F \cap H + F \cap B) + t\]  and sum; find \(t\) via inclusion–exclusion.
 

Answer 1

The Correct Option is A

Watching at least one: \(|F\cup H\cup B|=500-50=450.\) Using inclusion-exclusion with $t=|F\cap H\cap B|$: \[ 450=285+195+115-(45+70+50)+t=430+t \Rightarrow t=20. \] Only-F $=285-(70+45)+20=190$, 
Only-H $=195-(70+50)+20=95$, 
Only-B $=115-(45+50)+20=40$. 
Exactly one $=190+95+40=\boxed{325}$.